Solve: 4x+y=3 and 5x+4y=2?

Question

You don't have to solve it, but maybe explain me how that works.
Thank you

Answer 1

4x+y=3 ---equation 1
5x+4y=2 ---equation 2

from equation 1, make y the subject
y=3-4x

put this y in equation 2
5x+4(3-4x)=2
5x+12-16x=2
-11x=-10
x=10/11

put this x in y=3-4x
y=3-4(10/11)
y=3-40/11
y=-7/11

so the answer is x=10/11 and y= -7/11

Answer 2

One of the better methods to be comfortable with is to get used to is to solving for one variable and then use that to replace that value in the next equation. It is useful to be comfortable with it because it will allow you to solve for any number of unknowns given enough equations. For example... solve 5x+4y=2 for y 4y=2-5x subtract 5x fro both sides y=(2-5x)/4 divide both sides by 4 there that one is done and now that we know what y is equal to in terms of x we can take the next equation and wherever we see a y we just replace it with (2-5x)/4 4x+2y= -2 original equation 4x+2(2-5x)/4 = -2 with replacement, now we have a regular x= equation so solve for x 4x +((4 - 10x)/4) = -2 thats 4-10x over 4, now to add them we need a common denominator so... (16x + 4 -10x)/4 = -2 multiply both sides by 4 to get rid of the denominator 16x +4 -10x = -8 combine like terms and move the 4 6x = -12 divide both sides by 6 x= -2 now we have a numerical value for x Finally just plug that into the first equation and solve for y 5(-2) + 4y =2 multiply that out and move it to the other side remembering to change the sign 4y = 12 divide by 4 y = 3 there you go x= -2 and y=3 plug those values into your equations and see if they work out sorry if that was over simplified I never know if I'm writing to someone in the 3rd grade or high school anyway hope it helps good luck

Answer 3

we have to get one equation to say x= or y=
and then we can substitute into the other

let's take the first one
4x + y = 3...subtract 4x from both sides
y = 3 - 4x

now that we 'know' what Y is - substitute into the other equation

5x + 4(3-4x) = 2...simplify
5x + 12 - 16x = 2
12-11x = 2..add 11x to both sides
12=2+11x...subtract 2 from both
10 = 11X

now solve for X by dividing both sides by 11
and place back into our first equation we started with

then you have Y - and should work in BOTH equations

have a Happy Thanksgiving!

Answer 4

You need to isolate one of the variables and then substitute for it in the other equation. For example using equation 1 you can have y=3-4x. Now take equation 2 and change it to 5x+4(3-4x)=2. now you can solve for x in equation 2 and then plug this x value into equation 1.

Answer 5

from the first equation 4x+y=3
y=3-4x
Plug in this y in the other equation 5x+4y=2 for y.
solve for y.
go back to 4x+y=3 and solve for x.

Answer 6

4x + y = 3 - EQ 1

5x + 4y = 2 - EQ 2

Multiply EQ 1 by 4

16x + 4y = 12 - EQ 3

Now subtract EQ 2 from EQ 3

16x + 4y - 5x - 4y = 12 - 2

11x = 10
x = 10/11

Sub value back into EQ 1

4x + y = 3
40 / 11 + y = 3

y = -7/11

to check put in EQ 2.

5x + 4y = 2

50 / 11 + -28/11 = 2 (Hence it works)

Answer 7

{4x+y=3 {-16x-4y=-12
{5x+4y=2 {5x+4y=2 +
-11x=-10
x=10/11 so
4x+y=3
40/11 +y=3
y=-7/11

Answer 8

lots of math here so ill keep it simple

4x+y=3 solve for y
-4x =-4x
y=3-4x

Take this value for y and plug it into your other equation
5x+4(3-4x)=2
5x+12-16x=2
-11x=-10
divide by -11
x=10/11 or .91

Take this value for x and plug it into your answer for the first equation:

y= 3-4(.91)
y=3-3.64
y=-.64

So your answer for the following equation is this

Y= -.64
X= .91

Hope this helps

Answer 9

-16x - 4y = - 12
5x + 4y = 2-----ADD
-11x = - 10
x = 10 / 11

40/11 + y = 3
y = 33/11 - 40 / 11
y = - 7 / 11

x = 10 / 11 , y = - 7 / 11

Answer 10

you need to use subsitution method

4x+y=3 is the same as y=3-4x
subsitute that into the other equation so
5x +4(3-4x)=2
then solve for x, find x than solve for y